What is the tangent of y=e^{5x},\at x= 1/5 ln(5) ?

The tangent of y=e^{5x},\at x= 1/5 ln(5) is y=25x+5-5ln(5)

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tangent of y=e^{5x},\at x= 1/5 ln(5)

tangent of

y = e^{5 x}, at x = \frac{1}{5} ln (5)

y = 25 x + 5 - 5 ln (5)

Solution steps

Find the tangent point:

(\frac{1}{5} ln (5), 5)

Find the slope of

y = e^{5 x} : \frac{d y}{d x} = e^{5 x} \cdot 5

Compute the slope of

y = e^{5 x}

x = \frac{1}{5} ln (5) : m = 25

Find the line with slope m=

25

and passing through

(\frac{1}{5} ln (5), 5) : y = 25 x + 5 - 5 ln (5)

y = 25 x + 5 - 5 ln (5)

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\frac{d}{d x} (arccsc (4 x^{2} + 1))

\int (5 x^{4} + \frac{x}{7}) d x

\frac{d y}{d x} + y + 2 cos (x) = 0

\frac{d}{d x} (\frac{( x ^{2} + 3 )}{x})

What is the tangent of y=e^{5x},\at x= 1/5 ln(5) ?
The tangent of y=e^{5x},\at x= 1/5 ln(5) is y=25x+5-5ln(5)